Let's talk this out and see if we can work it out.
The sum of the first N odd integers means,
Where N is how many odd numbers we're adding.
Let's choose numbers for N, and see if we can find a pattern.
N=1 --> 1
(sum of the first odd integer)
N=2 --> 1 + 3 = 4
(sum of the first 2 odd integers)
N=3 --> 1 + 3 + 5 = 9
(sum of the first 3 odd integers)
N=4 --> 1 + 3 + 5 + 7 = 16
Do you notice a pattern yet? Take a look at when N = 2, what is the sum? That's right, 4!
and when N = 3... the sum is 9.
N = 4 the sum is 16....
I see a pattern, do you?
Answer: If you don't, you'll notice that the sum of the first N odd integers is always = N2
Sn = n^2
The formula for the sum of the squares of odd integers from 1 to n is n(n + 1)(n + 2) ÷ 6. EXAMPLE : Sum of odd integer squares from 1 to 15 = 15 x 16 x 17 ÷ 6 = 680
The arithmetic sequence of odd integers is 1, 3, 5, 7, 9, ... where the common difference is 2. The sum of the first thirty odd integers can be found by using the sum formula such as: Sn = n/2[2a1 + (n - 1)d], where a1 = 1, n = 30, and d = 2. So, S30 = (30/2)[(2)(1) + (30 - 1)(2)] = (15)[2 + (29)(2)] = (15)(60) = 900
If n is an odd integer then the next two consecutive odd integers are n+2 and n+4.
Sum of first n integers is n/2 times n + 1 ie 27.5 x 56 which is 1540
The sum of three consecutive odd integers, starting with N, is expressed as N + (N+2) + (N+4). If that sum is 363, then you have 3N + 6 = 363. Solve for N and you have 119.Since N (119) is odd, the answer and question are valid. (If N had been even, the question would have been invalid, and the answer would have been meaningless. This test is simply a sanity check.) The three numbers are 119, 121, and 123.
First of all, the sum of 3 odd numbers always equals an odd number and 176 is even so this is NOT possible.If you did not know this you could let n be the first odd number, n+2 is the next and n+4 is the third.Add them up and we have n+n+2+n+4=176 or3n+6=176 so3n=170Now you see that 170 is NOT divisible by 3 so ONCE again you see that there is NO solution to this problem. That is to say, there are NOT 3 odd consecutive integers whose sum is 176.
2n can be split into 2 n's so: n+n then add one to one of the n's and subtract one from one of the n's n+1+n-1 ^two consecutive odd integers^
let n = first odd number; n + 2 = 2nd odd number ,etc. n + n+2 + n+4 + n +6 = 4n + 12 = -72 4n = -84 n = -21 -21,-19,-17 and -15
False. let the integers be n, n+1 and n+2 3n+3 is there sum and we need this to be even for all integers n. if n is odd, then 3n is odd ( take n=5 3x5=15 odd) any odd number +3 is even. if n is even, then 3n is even and an even number plus and 3 is odd so the answer is false You could just say or prove it is false with a single counter example. Take the 3 consecutive integers, 2,3,4 their sum is 9 and you are done. I mentioned the 3n+3 so you can see why it is false for even set of 3 when the first of the 3 consecutive numbers is even.
Sn = n*(n+1)
n(n+2)(n+4) = 24
Call the first number n, and the next one n+1 Since their sum is 450 we have 2n+1=450 or 2n=449, but 449 is an odd number so it is not divisible by 2. This tells us there are no two such integers.
Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275
You can find that out by trial-and-error, that is, experiment with different numbers. Or, you can call the first number "n", in which case the second number is "n+2", and you solve the equation:n + (n+2) = 32
The sum of two even numbers is always an even number. The sum of two odd numbers is always an even number.The sum of an even number and an odd number is always an odd number.Because integers alternate between even and odd, adding two consecutive integers will always result the sum of an odd number and an even number, which, as stated above, will always be an odd number.Another way to look at it:An even number is divisible by 2, therefore, 2*n (where n is an integer) is always an even number.2*n+1 would be the next number after 2*n (so 2*n & 2*n+1 are consecutive integers).If we add these numbers together, we get 4*n+1 = 2(2*n)+1. Since 2(2*n) is an even number, adding 1 makes it an odd number._________________________________________________________________________I'm not going to but in or anything, seriously that was a good answer
If n is the smallest of the four integers, their sum is 4n+6.
The formula n*(n+1) is used to find the sum of n positive integers. Th sum of positive integers up to 500 can be calculated as 250*251=62,750.
The sum of the first n odd numbers is n squared. 25 x 25 = 625
The sum of nth odd number can be calculated by S(n+1) = (n+1)(n+2)/2
The sum of the first n positive integers (i.e. 1+2+3+...+n) is given by n(n+1)/2 so the sum of the first 25 positive integers is 1+2+3+...+25 = 25x26/2 = 325. The sum of the first 25 EVEN numbers will be double this i.e. 2+4+6+...+50 = 650.
55 Basically the sum of al integers from 1 to n is Sn=(n²+n)/2 (Think why is this number always an integer!)
(1 + 25) x 25/2 ie 425. The sum of the first n integers is (n + 1) x n/2